p07_015.pdf

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15. There is no acceleration, so the lifting force is equal to the weight of the object. We note that the

person’s pull

is equal (in magnitude) to the tension in the cord.

(a) As indicated in the

hint,

tension contributes twice to the lifting of the canister: 2T =

mg.

Since,

, we ﬁnd

= 98 N.

(b) To rise 0.020 m, two segments of the cord (see Fig. 7-28) must shorten by that amount. Thus, the

amount of string pulled down at the left end (this is the magnitude of

the downward displacement

of the hand) is

= 0.040 m.

and

are downward, then Eq. 7-7 leads to

= (98)(0.040) =

3.9 J.

(d) Since the force of gravity

(with magnitude

mg)

is opposite to the displacement

= 0.020 m

(up) of the canister, Eq. 7-7 leads to

−(196)(0.020)

−3.9

J. This is consistent with

Eq. 7-15 since there is no change in kinetic energy.